How you solve this? #lim_(n->oo)(|__x__|+|__3^2x__|+...+|__(2n-1)^2x__|)/n^3#

This can be understood as the realization of the Riemann-Stieltjes integral of

but

Let us first find a closed formula for:

The first few terms are:

Write down the sequence of differences between consecutive terms:

Write down the sequence of differences of those differences:

Write down the sequence of differences of those differences:

Note also that:

So:

So:

To solve the limit lim_(n->oo)(|x|+|3^2x|+...+|(2n-1)^2x|)/n^3, we can rewrite the expression as the sum of individual limits.

First, let's consider the limit of each term in the numerator.

For the term |x|, as n approaches infinity, x remains constant, so the limit is |x|.

For the term |3^2x|, the exponent 3^2x grows faster than n, so as n approaches infinity, this term becomes negligible and approaches 0.

Similarly, for the terms |(2n-1)^2x|, the exponents (2n-1)^2x also grow faster than n, so these terms also become negligible and approach 0 as n approaches infinity.

Therefore, the numerator simplifies to |x|.

The denominator, n^3, grows faster than all the terms in the numerator, so as n approaches infinity, the denominator dominates and the limit approaches 0.

In conclusion, the solution to the given limit is 0.

To solve the limit ( \lim_{n \to \infty} \frac{|x| + |3^2x| + \ldots + |(2n - 1)^2x|}{n^3} ), we first recognize that as ( n ) approaches infinity, the expression inside the absolute values will keep growing larger since each term is squared. Therefore, we can simplify the absolute values as follows:

[ |(2k - 1)^2x| = (2k - 1)^2x ]

Where ( k ) represents the terms in the summation.

Now, let's rewrite the given expression:

[ \lim_{n \to \infty} \frac{|x| + |3^2x| + \ldots + |(2n - 1)^2x|}{n^3} ] [ = \lim_{n \to \infty} \frac{x + 3^2x + \ldots + (2n - 1)^2x}{n^3} ] [ = \lim_{n \to \infty} \frac{x(1^2 + 3^2 + \ldots + (2n - 1)^2)}{n^3} ]

Now, we recognize that the sum ( 1^2 + 3^2 + \ldots + (2n - 1)^2 ) is equivalent to the sum of the squares of the odd integers up to ( 2n - 1 ), which can be expressed as a known formula:

[ 1^2 + 3^2 + \ldots + (2n - 1)^2 = \frac{n(2n - 1)(2n + 1)}{3} ]

Substitute this into the expression:

[ \lim_{n \to \infty} \frac{x \cdot \frac{n(2n - 1)(2n + 1)}{3}}{n^3} ] [ = \lim_{n \to \infty} \frac{x(2n^3 + \ldots)}{3n^3} ] [ = \lim_{n \to \infty} \frac{x}{\frac{3}{2n} + \ldots} ]

As ( n ) approaches infinity, the terms with ( n ) in the denominator become negligible, leaving us with:

[ \lim_{n \to \infty} \frac{x}{\frac{3}{2n}} = \lim_{n \to \infty} \frac{2nx}{3} ]

Therefore, the limit evaluates to ( \frac{2x}{3} ).